Question

4. A geometric sequence is defined recursively by a(1) = 40 and a(n)= a(n-1)

(a) Write out the first four terms of this sequence.

(b) Is the 9th term of this sequence larger or smaller than 1/10? Show the calculation that you use to determine your answer.

Answer 1

The complete question is:

A geometric sequence is defined recursively by

    $a_{(1)}=40\text{ and }a_n=(-1/2)a_{(n-1)}$

(a) Write out the first four terms of this sequence.

(b) Is the 9th term of this sequence larger or smaller than 1/10? Show the calculation that you use to determine your answer.

Answer:

• The 9th term is larger than 1/10.

Explanation:

You are given the first term,    $a_(1)=40$ , and the recursive formula

   $a_{(n)}=(-1/2)a_{(n-1}$

Thus, you can find the sequence of terms by multiplying each term by the constant ratio, (-1/2).

(a) First four terms of the sequence

$a_{(1)}=40\\ \\ a_{(2)}=40\times (-1/2)=-20\\ \\ a_{(3)}=-20\times (-1/2)=10\\\\ a_{(4)}=10\times (-1/2)=-5$

Those are the first four terms.

(b) 9th term

You can write the explicit formula of the sequence as:

$a_{(n)}=a_{(1)}\times r^{(n-1)}\\ \\ a_{(n)}=40(-1/2)^{(n-1)}$

Thus, for the 9th term, n = 9, and the term is:

$a_{(9)}=40\times (-1/2)^{(9-1)}\\ \\ a_{(9)}=40\times(-1/2)^8\\ \\ a_{(9)}=40\times(1/256)=0.1526$

Since 1/10 = 0.1, and 0.1 < 0.1526, the 9th term is larger than 1/10.